SOLUTION: Find the value of k the system has infinite number of solution x+(k+1)y=5;(k+1)x+9y=8k-1

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Question 1085977: Find the value of k the system has infinite number of solution x+(k+1)y=5;(k+1)x+9y=8k-1
Found 2 solutions by ikleyn, dubeypriyanshu436@gmail.com:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
1.  One way to solve it is using determinant.


2.  The other way is  THIS:

    the system has INFINITELY MANY solutions if and only if the two equations are equivalent.

    In other words, if and only if

    1%2F%28k%2B1%29 = %28k%2B1%29%2F9 = 5%2F%288k-1%29.    (1)

    (the coefficients and the right side terms are proportional with the same proportionality coefficient).


3.   From  1%2F%28k%2B1%29 = %28k%2B1%29%2F9  you have  %28k%2B1%29%5E2 = 9  ====>  k+1 = +/-3  ====>  a) k = -1 + 3 = 2,  and  b) k = -1 - 3 = -4.



     Now check k = 2 for this proportion: %28k%2B1%29%2F9 = 5%2F%288k-1%29.

     You have %282%2B1%29%2F9 = 1%2F3 (left side)  and  5%2F%288%2A2-1%29 = 5%2F15 = 1%2F3  (right side).

     So, the value k = 2 satisfies (1).



     Now check k = -4 for this proportion: %28k%2B1%29%2F9 = 5%2F%288k-1%29.

     You have %28-4%2B1%29%2F9 = -1%2F3 (left side)  and  5%2F%288%2A%28-4%29-1%29 = 5%2F%28-33%29  (right side).

     So, the value k = -4 does not satisfy (1).

Answer. The value of "k" under the question is k = 2.


See the lesson
    - Geometric interpretation of the linear system of two equations in two unknowns
in this site.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Systems of two linear equations in two unknowns".



Answer by dubeypriyanshu436@gmail.com(1) About Me  (Show Source):
You can put this solution on YOUR website!
1/k+1=k+1/9=5/8k-1
K+1×k+1=9
(K+1)^2=9

K+1=√9

K+1=+-3

Take k+1=3
K=3-1
K=2

K=2. Ans

Take k+1=-3
K=-3-1
K=-4
Now we can check our and answer
Putk=2
x+2+1×y=5;2+1×x+9y=8×2-1
x+3y=5;3x+9y=15
x=5-3y;3×5-3y+9y=15
15+6y=15
y=0
Put y=0
x=5-3×0
x=5
Put k=-4
x-3y=5
x=5+3y
Put the value of x in equation
5+3y-3y=5
5=5
Put in 2equation
is not equal Lhs is not equal to Rhs
So the and is k=2